intersection of a line and a plane

Question

My kid was given this problem:

Line $\ell_1$ is given by $(1,4,2)+t(2,-1,-2)$. Line $\ell_2$ passes through the point $A=(6,0,-3)$, is parallel to the plane $\pi$ given by $3x-4y+z-10=0$, and intersects $\ell_1$ at $B$. (1) Find, in terms of $t$, the directional vector of $\ell_2$. (2) Find $B$.

So I decided to find $B$ first. I found the plane parallel to $\pi$ through $A$ to be $3x-4y+z-15=0$, and put $(1-2t,4-t,2-2t)$ in for $(x,y,z)$ in that equation to find $B$. I wound up with $(7.5,0.75,-1.5)$, which seems exceedingly unlikely for a normal high-school math-book problem (and subsequent analysis shows, in fact, an inconsistency). What did I do wrong? What should I have done instead?

Answer 1

The normal vector of the given plane $3x-4y+z-10=0$ is $(3, -4,1)$. Then, the equation of the plane parallel to it and passing the point $A(6,0,-3)$ is

$$(x-6, y, z+3)\cdot (3, -4,1)=0\implies 3x-4y+z-15=0$$

Substitute the given line $(x,y,z)=(1+2t, 4-t, 2-2t)$ into above plane to get $t$ for point B,

$$3(1+2t)-4(4-t)+(2-2t)-15=0\implies t=\frac{13}4$$

Plug $t$ back into the given line $(1+2t, 4-t, 2-2t) $ to obtain the point B$(\frac{15}2,\frac34,-\frac92)$.

Answer 2

You made at least one arithmetic mistake. Everything else about your solution was fine.

First, you are correct that $A$ and $B$ will be contained in the plane given by the equation $3x-4y+z=15$. Let's call this plane $\pi_2$.

Next, you tried to find $B$ by using the fact that $B$ is the unique point in $\ell_1\cap\pi_2$. Plugging $(1+2t,4-t,2-2t)$ into $3x-4y+z=15$, you get that $t=\frac{13}{4}=3.25$. Hence $B=(7.5,0.75,-4.5)$.

Now $\ell_2$ is the set of points of the form $A+t(B-A)=(6,0,-3)+t\cdot(1.5,0.75,-1.5)$. So the directional vector is $(1.5,0.75,-1.5)$.

You can check that every point of the form $A+t(B-A)$ is on $\pi_2$.